Playframework:[RuntimeException:java.lang.reflect.InvocationTargetException]

Question

我正在尝试基于Zentask示例创建一个简单的登录 - zentask - playframework,但是当我单击调用Application.authenticate操作的登录按钮时,它会给出运行时异常.我用 - 错误标记了这一行

[RuntimeException: java.lang.reflect.InvocationTargetException]

Application.java

public class Application extends Controller {

.........

public static class Login 
{
    public String email;
    public String password;

    public String validate() 
    {
        if (User.authenticate(email, password) == null) {
          return "Invalid user or password";
        }
        return null;
    }
}

   public static Result authenticate() 
    {
        Form<Login> loginForm = form(Login.class).bindFromRequest();  //--- error
        if(loginForm.hasErrors()) {
            return badRequest(login.render(loginForm));
        } else {
            session("email", loginForm.get().email);
            return redirect(
                routes.Application.index()
            );
        }
    }
}

我知道它与Login Class中的validate函数有关,因为当我在validate函数中删除对User.authenticate的调用时,它可以正常工作.但我无法弄明白.

用户类如下 -

@Entity
public class User extends Model
{
    @Id
    @Constraints.Required
    @Formats.NonEmpty
    public String userId;

    @OneToOne(cascade=CascadeType.PERSIST)
    AccountDetails accDetails;


    public static Model.Finder<String,User> find = new Model.Finder<String,User>(String.class, User.class);



    // Authenticate the user details
    public static User authenticate(String email, String password) 
    {
        String tempId = AccountDetails.authenticate(email, password).userId;

        return find.ref(tempId);
    }

    .. . . . . . .

}

和AccountDetails类 -

@Entity
public class AccountDetails extends Model
{
    @Id
    String userId;

    @Constraints.Required
    String emailId;

    @Constraints.Required
    String password;

    public static Model.Finder<String,AccountDetails> find = 
            new Model.Finder<String,AccountDetails>(String.class, AccountDetails.class);


    public static AccountDetails authenticate(String email, String password) 
    {       
        return find.where()
            .eq("email", email)
            .eq("password", password)
            .findUnique();
    }

}

Answer 1

我必须假设很多，但如果这就是你的堆栈跟踪的样子：

play.api.Application$$anon$1: Execution exception[[RuntimeException: java.lang.reflect.InvocationTargetException]]
                at play.api.Application$class.handleError(Application.scala:293) ~[play_2.10.jar:2.2.3]
                at play.api.DefaultApplication.handleError(Application.scala:399) [play_2.10.jar:2.2.3]
                at play.core.server.netty.PlayDefaultUpstreamHandler$$anonfun$3$$anonfun$applyOrElse$3.apply(PlayDefaultUpstreamHandler.scala:264) [play_2.10.jar:2.2.3]
                at play.core.server.netty.PlayDefaultUpstreamHandler$$anonfun$3$$anonfun$applyOrElse$3.apply(PlayDefaultUpstreamHandler.scala:264) [play_2.10.jar:2.2.3]
                at scala.Option.map(Option.scala:145) [scala-library.jar:na]
                at play.core.server.netty.PlayDefaultUpstreamHandler$$anonfun$3.applyOrElse(PlayDefaultUpstreamHandler.scala:264) [play_2.10.jar:2.2.3]
        ...
        Caused by: java.lang.NullPointerException: null
                at models.User.authenticate(User.java:26) ~[na:na]
                at controllers.Application$Login.validate(Application.java:50) ~[na:na]
                at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method) ~[na:1.8.0_05]
                at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62) ~[na:1.8.0_05]
                at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43) ~[na:1.8.0_05]
                at java.lang.reflect.Method.invoke(Method.java:483) ~[na:1.8.0_05]

那么问题的原因实际上是由 User 类中的 NPE 暗示的。如果您输入虚假凭据，您的查找器将找不到任何内容，并在 AccountDetails.authenticate() 中返回 null。

因此，在下面的方法中，您不检查 null 并尝试获取 userId，这会导致 NPE：

public static User authenticate(String email, String password) {
  String tempId = AccountDetails.authenticate(email, password).userId;

  return find.ref(tempId);
}

如果你只是正确检查 null ，你将获得所需的功能：

public static User authenticate(String email, String password) {
    User user = null;

    AccountDetails accountDetails = AccountDetails.authenticate(email, password);
    if (accountDetails != null) {
        user = find.ref(accountDetails.userId);
    }

    return user;
}